Problem 115 - A Complete Resource Book in Chemistry for JEE Main

Question

A \(0.2\) molal aqueous solution of a weak acid (HX) is \(20 \%\) ionized. The freezing point of this solution is (Given \(\mathrm{K}_{\mathrm{f}}=1.86^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}\) for water \()\) (a) \(-0.45^{\circ} \mathrm{C}\) (b) \(-0.90^{\circ} \mathrm{C}\) (c) \(-0.21^{\circ} \mathrm{C}\) (d) \(-0.43^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

The freezing point is approximately \(-0.45\, ^\circ C\), which corresponds to option (a).

Step 1: Understanding the problem

We have a weak acid (HX) which is 20% ionized in a 0.2 molal solution. We are to find the freezing point depression using the given cryoscopic constant of water, \( K_f = 1.86\, ^\circ C\, kg\, mol^{-1} \).

Step 2: Calculate van't Hoff factor

The weak acid HX is 20% ionized, meaning 20% of HX dissociates into \( H^+ \) and \( X^- \). For a weak acid, the van't Hoff factor, \( i \), can be calculated as \( i = 1 + \alpha \), where \( \alpha \) is the degree of ionization. Here, \( \alpha = 0.20 \), so \( i = 1 + 0.20 = 1.20 \).

Step 3: Apply freezing point depression formula

The freezing point depression, \( \Delta T_f \), can be calculated using the formula \( \Delta T_f = i \times K_f \times m \), where \( i \) is the van't Hoff factor, \( K_f \) is the cryoscopic constant, and \( m \) is the molality. Substituting the known values, \( \Delta T_f = 1.20 \times 1.86 \times 0.2 \).

Step 4: Calculate the freezing point depression

Perform the calculation: \( \Delta T_f = 1.20 \times 1.86 \times 0.2 = 0.4464\, ^\circ C \).

Step 5: Determine the new freezing point

The normal freezing point of pure water is \( 0\, ^\circ C \). The solution's freezing point is lowered by the depression amount: therefore, the new freezing point is \( 0 - 0.4464 \approx -0.45\, ^\circ C \).

Step 6: Choose the correct option

Comparing \( -0.45\, ^\circ C \) with the given options, option (a) \(-0.45\, ^\circ C\) is the correct answer.

Key Concepts

Understanding the van't Hoff Factor

The van't Hoff factor, often denoted by the symbol \(i\), is crucial in understanding colligative properties like freezing point depression. It describes the number of particles a compound generates upon dissolving. For strong electrolytes, which completely dissociate in solution, \(i\) equals the total number of ions produced. However, for weak electrolytes like weak acids, \(i\) can be less straightforward.
A weak acid, such as HX, only partially ionizes in solution. In this exercise, HX \(\rightarrow H^+ + X^-\), and the problem states that it is 20% ionized. This translates to a 20% conversion of the initial HX to \(H^+\) and \(X^-\).
You calculate the van't Hoff factor for a weak electrolyte using the formula:

• \(i = 1 + \alpha\),

where \(\alpha\) is the degree of ionization. For our weak acid HX:

• \(\alpha = 0.20\), so \(i = 1 + 0.20 = 1.20\).

This calculation indicates that each mole of HX results in 1.20 moles of particles in the solution. This correction helps accurately predict the freezing point depression.

Exploring the Cryoscopic Constant

The cryoscopic constant, \(K_f\), is a property specific to each solvent, indicating how much the freezing point will lower per molal concentration of a non-volatile solute. It represents the solvent's response to solutes, making it an essential component in calculating freezing point depression.
For water, this constant has been experimentally determined as \(1.86\, ^\circ C\, kg\, mol^{-1}\). This means that for every molal of solute particles added to water, the freezing point is expected to drop by \(1.86^\circ C\).
It’s important to use the cryoscopic constant in conjunction with the van't Hoff factor and the molality to calculate the change in freezing point efficiently:

• \(\Delta T_f = i \times K_f \times m\),

where \(i\) is the van't Hoff factor, \(K_f\) is the cryoscopic constant, and \(m\) is the molality of the solution. By understanding and applying this constant correctly, students can accurately determine how solutions behave under increased solute concentration.

Understanding Weak Acid Ionization

Weak acid ionization involves partially dissociating into ions in a solution, meaning not all the acid molecules present dissociate to form hydrogen ions. This level of ionization can be represented by the degree, \(\alpha\), which can numerically describe what fraction of the original acid molecules ionize.
For weak acids like HX:

• \(\alpha = 0.20\) implies that 20% of HX dissociates into \(H^+\) and \(X^-\) ions.

This partial ionization leads to fewer free ions in solution compared to a strong acid of the same concentration, affecting calculations for properties like freezing point depression.
Knowing the degree of ionization is crucial when determining the van't Hoff factor, as it affects the total particle count in the solution, which in turn impacts colligative properties. Understanding the concept of weak acid ionization helps properly predict and manipulate the solutions' physical properties in varied chemical scenarios.